Q1:jipmat 2025QA › Time & WorkMediumQA · MCQ

40 men can complete a work in 48 days. 64 men started and did the same work for x days. After x days, 32 men increased, so, the remaining work is completed in $16\frac{2}{3}$ days, then the value of x is

• A5
• B8
• C10
• D6

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The Setup: We are measuring this task in 'man-days' (the amount of grinding one guy does in one day). The total work is a fixed pool of points, and we just need to track who did how much. Step 1: Calculate the total work pool. $$\text{Total Work}=40\text{ men} \times 48\text{ days}=1920\text{ man-days}$$ Step 2: Track the grinding phases. Phase 1: $64\text{ men}$ work for $x\text{ days}$. Work done in Phase 1 $=64x$. Phase 2: The squad increases by $32$ men ($64+32=96\text{ men}$). They grind for $16\frac{2}{3}$ days. Convert $16\frac{2}{3}$ to an improper fraction: $\frac{50}{3}$. Work done in Phase 2 $=96 \times \frac{50}{3}$. $96 \div 3=32$, so Phase 2 work $=32 \times 50=1600\text{ man-days}$. Step 3: Set up the master equation. Total work must equal the sum of both phases. $$64x+1600=1920$$ $$64x=1920-1600$$ $$64x=320$$ $$x=\frac{320}{64}=5$$ Final Answer: 5